Let $S$ be the sphere centered at $(-2, 4, 1)$ with radius $3$. What is the triple integral of the scalar field $f(x, y, z)$ over $S$ in spherical coordinates? Assume that $x$, $y$, and $z$ are expressed in terms of $\rho$, $\theta$, and $\varphi$. Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^\pi \int_0^{2\pi} \int_0^3 f(x - 2, y - 4, z + 1) \rho\sin^2(\varphi) \, d\rho \, d\theta \, d\varphi$ (Choice B) B $ \int_0^\pi \int_0^{2\pi} \int_0^3 f(x + 2, y - 4, z - 1) \rho^2\sin(\varphi) \, d\rho \, d\theta \, d\varphi$ (Choice C) C $ \int_0^\pi \int_0^{2\pi} \int_0^3 f(x + 2, y - 4, z - 1) \rho\sin^2(\varphi) \, d\rho \, d\theta \, d\varphi$ (Choice D) D $ \int_0^\pi \int_0^{2\pi} \int_0^3 f(x - 2, y + 4, z + 1) \rho^2\sin(\varphi) \, d\rho \, d\theta \, d\varphi$
The bounds we'll use are $0 < \theta < 2\pi$ and $0 < \varphi < \pi$. Here is the change of variables for spherical coordinates. $\begin{aligned} x &= \rho \cos(\theta) \sin(\varphi) \\ \\ y &= \rho \sin(\theta) \sin(\varphi) \\ \\ z &= \rho \cos(\varphi) \end{aligned}$ We want to represent the sphere $S$ with bounds in spherical coordinates. The standard unit sphere needs $\varphi$ to range across $[0, \pi]$, $\theta$ to range across $[0, 2\pi]$, and $\rho$ to range across $[0, 1]$. The sphere $S$ has radius $3$, so it needs $\rho$ to range across $[0, 3]$ : $ \int_0^\pi \int_0^{2\pi} \int_0^3 \cdots \, d\rho \, d\theta \, d\varphi$ Although $S$ is a unit sphere, it's also shifted so that it's centered at $(-2, 4, 1)$. How can we can keep the bounds for the sphere and integrate over $S$ ? We can subtract $2$ to $x$ everywhere to transform $S$ two units to the left. We write $f(x - 2, y, z)$ to account for this shift. This kind of a shift has no effect with change of variables, because its Jacobian is $1$. We can do the same for $y$ and $z$. So our integral looks like this now: $ \int_0^\pi \int_0^{2\pi} \int_0^3 f(x - 2, y + 4, z + 1) \cdots \, d\rho \, d\theta \, d\varphi$ The final step is finding the Jacobian of spherical coordinates, which we'll need to multiply in to get the final integral. $J(\rho, \theta, \varphi) = \rho^2\sin(\varphi)$ [Derivation] The integral in spherical coordinates: $ \int_0^\pi \int_0^{2\pi} \int_0^3 f(x - 2, y + 4, z + 1) \rho^2\sin(\varphi) \, d\rho \, d\theta \, d\varphi$